Let \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ n(n+1) } =\frac { n }{ (n+1) } \)
For n =1
 P(1) = \(\frac { 1 }{ 1(1+1) } =\frac { 1 }{ 1+1 } \Rightarrow \frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
∴ P(1) is true
Let P(n) be true for n = k
∴  P(k) = \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ k(k+1) } =\frac { k }{ (k+1) } \)   ...(i)
For n = k+1
∴ P(k+1)= \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ k(k+1) } +\frac { 1 }{ (k+1)(k+2) } =\frac { k+1 }{ k+2 } \)
= \(\frac { k }{ k+1 } +\frac { 1 }{ (k+1)(k+2) } =\frac { k(k+2)+1 }{ (k+1)(k+2) } =\frac { { k }^{ 2 }+2k+1 }{ (k+1)(k+2) } \)       [Using (i)]
=  \(\frac { (k+1)^{ 2 } }{ (k+1)(k+2) } =\frac { k+1 }{ k+2 } \)= P(k+1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all n∊N.

Let P(n) = (2n + 7) < (n + 3)2
For n =1
P(1) =  (2\(\times\)1+7)<(1+3)²
⇒ 9 < 16
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = (2k+7)<(k+3)²    ...(i)
For n = k+1
P(k+1) =2(k+1)+7<(k+1+3)²
⇒ 2(k+1)+7<(k+4)²​        From (i)​
2k+7<(k+3)²​​​​​
Adding 2 on both sides
2k+7+2<(k+3)²+2
⇒  2(k+1)+72+9+6k+2
⇒ 2(k+1)+72+6k+112+8k+16
⇒ 2(k+1)+7<(k+4)²
∴ P(k +1) is true
Thus P(k) is true
⇒ P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all n∊N.

Let P(n) = x²ⁿ -y²ⁿ is divisible by (x + y)
For n =1
P(1) x^2\(\times\)1-y^2\(\times\)1 is divisible by (x+y)
⇒ (x+y)(x-y) is divisible by (x+y)
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = x^2k -y^2k is divisible by (x + y)
⇒ x^2k-y^2k = λ(x+y)    ... (i)
For  n = k+1
P(k+1)= x^2(k+1)-y^2(k+1) isdivisible by (x+y)
Now x^2k+2-y^2k+2
= x^2k+2-x^2ky²+x^2ky²-y^2ky²
= x^2k.x²-x²y²+x^2ky²-y^2k.y²
= x^2k(x²-y²)+y²(x^2k-y^2k)
=x^2k(x²-y²)+y2.λ(x+y)   [Using (i)]
= (x+y) [x^2k(x-y)+λy²]
⇒ x^2k+2-y^2k+2 is divisible by (x+y)
∴ P(k+1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction,P(n) is true for all n ∊ N.

Let P(n) =1+2+3+....+\(n<\frac { 1 }{ 8 } (2n+1)^{ 2 }\)
For n =1
P(1) = \(1<\frac { 1 }{ 8 } (2\times +1+)^{ 2 }\Rightarrow 1<\frac { 9 }{ 8 } \)
∴ P(1) is true
Let p(n) be true for n = k
∴  P(k) = 1 + 2 + 3 + .....+\(k<\frac { 1 }{ 8 } (2k+1)^{ 2 }\) ..(i)
For n = k+1
P(k+1) =1+2+3+...+k+(k+1)<\(\frac { 1 }{ 8 } \) (2k+3)²
From (i) we have
1+2+3+.....+ k<\(\frac { 1 }{ 8 } \) (2k+3)²
Adding (k + 1) on both sides, we get
1+2+3+...... +k+(k+1)<\(\frac { 1 }{ 8 } \)(2k+1)²+(k+1)
1+2+3+...... +k+(k+1)<\(\frac { 1 }{ 8 } \) [4k²+4k+1+8k+8]
1+2+3+...... +k+(k+1) <\(\frac { 1 }{ 8 } \)[4k²+12k+9]
1+2+3+...... +k+(k+1)<\(\frac { 1 }{ 8 } \)(2k+3)²
  \(\frac { k+1 }{ 2 } \)+[2\(\times\)1+(k+1-1)1]<\(\frac { 1 }{ 8 } \)(2k+3)²
⇒ \(\frac { (k+1)(k+2) }{ 2 } \)<\(\frac { 1 }{ 8 } \)(2k+3)²
⇒ 4(k + 1)(k+2)<4k² + 9 + 6k
⇒ 4(k² +3k+2)<4k² + 9 + 6k
⇒ 4k²+12k+8<4k²+9+6k
⇒  8 < 9
∴  P(k + 1) is true.
Thus P(k) is true ⇒ P(k + 1) is true.
Hence by principle of mathematical induction,p(n) is true for all n ∊ N.

Let P(n)=  1²+3²+5²+....+(2n-1)² = \(\frac { n(2n-1)(2n+1) }{ 3 } \)
For n =1
P(1) = (2\(\times\)1-1)²
= \(\frac { 1(2\times 1-1)(2\times 1+1) }{ 3 } \)
⇒ 1= \(\frac { 1\times 1\times 3 }{ 3 } \)
∴ P(1) is true
Let P(n) be true for n = k
∴  P(k)= 1²+3²+5²+...+(2k-1)² =  \(\frac { k(2k-1)(2k+1) }{ 3 } \)...(i)
For n = k+1
R.H.S =\(\frac { (k+1)(2k+1)(2k+3) }{ 3 } \)
L.H.S =\(\frac { k(2k-1)(2k+1) }{ 3 } +(2k+1)^{ 2 }\) +(2k+1)²    [using(i)]
= (2k+1) \(\left[ \frac { k(2k-1) }{ 3 } +(2k+1) \right] \)
= (2k+1) \(\left[ \frac { 2{ k }^{ 2 }-k+6k+3 }{ 3 } \right] \)
= \(\frac { (2k+1)(2{ k }^{ 2 }+5k+3) }{ 3 } \)
= \(\frac { (2k+1)(k+1)(2k+3) }{ 3 } \)
= \(\frac { (k+1)(2k+1)(2k+3) }{ 3 } \)
∴  P(k+1) is true
Thus P(k) is true ⇒ P(k+1) is true
Hence by principle of mathematical induction, P(n) is true for all n ∊  N.

Let P(n) = :a+ar+ar²+.......+ar^n-1 = \(\frac { a({ r }^{ n }-1) }{ r-1 } \)
For n =1
P(1) = ar^n-1 = \(\frac { a({ r }^{ 1 }-1) }{ r-1 } \)
⇒ a =a
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = a+ar+ar²+.....+ar^k-1
= \(\frac { a({ r }^{ k }-1) }{ r-1 } \)...(i)
For n = k+1
R.H.S = \(\frac { a({ r }^{ k+1 }-1) }{ r-1 } \)
L.H.S =\(\frac { a({ r }^{ n }-1) }{ r-1 } \)+αr^k [Using (i)]
=  \(\frac { { ar }^{ k } }{ r-1 } -\frac { a }{ r-1 } +{ ar }^{ k }\)
=  \({ ar }^{ k }\left( \frac { 1 }{ r-1 } +1 \right) -\frac { a }{ r-1 } \)
= \({ ar }^{ k }\left( \frac { r }{ r-1 } \right) -\frac { a }{ r-1 } \)
= \(\frac { { ar }^{ k+1 } }{ r-1 } -\frac { a }{ r-1 } =\frac { { ar }^{ k+1 }-a }{ r-1 } \) 
= \(\frac { { a(r }^{ k+1 }-1) }{ r-1 } \)
∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all n ∊ N

Let P(n)= \(\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +....+\frac { 1 }{ 2^{ n } } =1-\frac { 1 }{ 2^{ n } } \)
For n =1
P(1)=\(\frac { 1 }{ 2^{ 1 } } =1-\frac { 1 }{ 2^{ 1 } } \Rightarrow \frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
∴ P(1) is true
Let P(n) be true for n = k
∴  P(k) =\(\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +....+\frac { 1 }{ 2^{ k } } \)
=  \(1-\frac { 1 }{ 2^{ k } } \)  ........(i)
For n = k + 1
P(k+1) = \(\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +....+\frac { 1 }{ 2^{ k } } +\frac { 1 }{ 2^{ k+1 } } \)
= \(1-\frac { 1 }{ 2^{ k+1 } } \)
=  \(1-\frac { 1 }{ 2^{ k } } +\frac { 1 }{ 2^{ k+1 } } \)    [Using (i)]
= 1- \(\left( \frac { 1 }{ 2^{ k } } -\frac { 1 }{ 2^{ k+1 } } \right) \)
=1-\(\left( \frac { 2-1 }{ 2^{ k+1 } } \right) =1-\frac { 1 }{ 2^{ k+1 } } \)
∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all n ∊ N.

Let P(n) = 1.3+3.5+5.7+....+(2n-1)(2n+1) = \(\frac { n(4{ n }^{ 2 }+6n-1) }{ 3 } \)
For n =1 
P(1) = (2\(\times\)1-1)(2\(\times\)1+1) = \(\frac { 1[4(1)^{ 2 }+6\times 1-1] }{ 3 } \)
\(1\times 3=\frac { 9 }{ 3 } \)⇒ 3 =3
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = 1.3 + 3.5 + 5.7+...+ (2k - 1) (2k + 1)
= \(\frac { k(4{ k }^{ 2 }+6k-1) }{ 3 } \)  ..(i)
For n = k + 1
P(k + 1)= 1.3 + 3.5 + 5.7+....+ (2k - 1)
 (2k+1)+[2(k+1)-1][2(k+1)+1]
∴ P(k+1)=\(\frac { k(4{ k }^{ 2 }+6k-1) }{ 3 } \)+(2k+1)(2k+3)
= \(\frac { { 4k }^{ 3 }+{ 6k }^{ 2 }-k+3({ 4k }^{ 2 }+8k+3) }{ 3 } \)
= \(\frac { { 4k }^{ 3 }+{ 6k }^{ 2 }-k+12{ k }^{ 2 }+24k+9 }{ 3 } \)
= \(\frac { { 4k }^{ 3 }+{ 18k }^{ 2 }+23k+9 }{ 3 } \)
= \(\frac { (k+1)({ 4k }^{ 2 }+14k+9) }{ 3 } \)
∴ P (k+1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all n ∊ N

Let P(n) =1.2.3+2.3.4+...+n(n+1)(n+2)= \(\frac { n(n+1)(n+2)(n+3) }{ 4 } \)
For n=1
P(1)=1\(\times\)2\(\times\)3 =\(\frac { 1\times 2\times 3\times 4 }{ 4 } \)
⇒ 6=6
∴ P(1) is true
Let P(n) be true for n = k.
∴   P(k) = 1.2.3+ 2.3.4 +...+ k(k + 1) (k + 2)
=\(\frac { k(k+1)(k+2)(k+3) }{ 4 } \)   ...(i)
For n = k + 1
p(k+1) = 1·2·3+ 2·3·4 + ..... + k(k + 1) (k + 2) + (k + 1) (k + 2) (k + 3)
= \(\frac { k(k+1)(k+2)(k+3) }{ 4 } \) + (k + 1) (k + 2) [Using (i)]
= (k + 1) (k + 2) (k + 3) \(\left[ \frac { k }{ 4 } +1 \right] \)
= (k + 1) (k + 2) (k + 3) \(\left[ \frac { k+1 }{ 4 } \right] \)
= \(\frac { (k+1)(k+2)(k+3)(k+4) }{ 4 } \)
∴   P(k + 1) is true.
Thus P(k) is true ⇒ P(k +1) is true
Hence by principle of mathematical induction, P(n) is true for all n ∊ N

Let
P(n)= 1+3+3²+.....+3^n-1 = \(\frac { ({ 3 }^{ n }-1) }{ 2 } \)
For n=1
P(1) = 1= \(\frac { { 3 }^{ 1 }-1 }{ 2 } \) ⇒1=1
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = 1+3+3²+...+3^k-1
=  \(\frac { ({ 3 }^{ k }-1) }{ 2 } \)   ....(i)
P(k) = 1 + 3 + 3² + .... + 3^k-1
 =   \(\frac { ({ 3 }^{ k }-1) }{ 2 } \)
For n= k+1
P(k+1)=1+3+3²+ .... +3^k-1+3k
=  \(\frac { ({ 3 }^{ k }-1) }{ 2 } +3^{ k }\)  [using (i)]
= \(\frac { { 3 }^{ k }-1+2.{ 3 }^{ k } }{ 2 } \)
= \(\frac { (3.3^{ k }-1) }{ 2 } =\frac { (3^{ k+1 }-1) }{ 2 } \)
∴ P(k + 1) is true.
Thus P(k) is true ⇒ P(k + 1) is true.
Hence by principle of mathematical induction, P(n) is true for all n ∊ N.